Is cos(2θ)=cos^2(θ)-sin^2(θ) ?

The answer to whether cos(2θ)=cos^2(θ)-sin^2(θ) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

prove cos(2θ)=cos^2(θ)-sin^2(θ)

prove

cos (2 θ) = cos^{2} (θ) - sin^{2} (θ)

True

Solution steps

cos (2 θ) = cos^{2} (θ) - sin^{2} (θ)

Manipulating left side

cos (2 θ)

Rewrite using trig identities

cos (2 θ)

Rewrite as

= cos (θ + θ)

Use the Angle Sum identity:

cos (s + s) = cos (s) cos (s) - sin (s) sin (s)

= cos (θ) cos (θ) - sin (θ) sin (θ)

= cos (θ) cos (θ) - sin (θ) sin (θ)

Simplify

cos (θ) cos (θ) - sin (θ) sin (θ) : cos^{2} (θ) - sin^{2} (θ)

cos (θ) cos (θ) - sin (θ) sin (θ)

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= cos^{2} (θ) - sin^{2} (θ)

= cos^{2} (θ) - sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e sin (3 θ) = 3 sin (θ) - 4 sin^{3} (θ)

p ro v e sin^{4} (t) - cos^{4} (t) = 1 - 2 cos^{2} (t)

p ro v e (1 - cos^{2} (x)) (1 + cot^{2} (x)) = 1

p ro v e \frac{1 - cos ( - x )}{sec ( - x ) - 1} = cos (x)

p ro v e cos (3 θ) = 4 cos^{3} (θ) - 3 cos (θ)

Is cos(2θ)=cos^2(θ)-sin^2(θ) ?
The answer to whether cos(2θ)=cos^2(θ)-sin^2(θ) is True